Question: Simplify and expand the following expression: $ \dfrac{2r + 8}{r + 4}-\dfrac{r + 9}{2r + 3} $
In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r + 4)(2r + 3)$ Multiply the first term by $\dfrac{2r + 3}{2r + 3}$ $ \begin{align*} \dfrac{2r + 8}{r + 4} \times \dfrac{2r + 3}{2r + 3} & = \dfrac{(2r + 8)(2r + 3)}{(r + 4)(2r + 3)} \\ & = \dfrac{4r^2 + 22r + 24}{(r + 4)(2r + 3)}\end{align*} $ Multiply the second term by $\dfrac{r + 4}{r + 4}$ $ \begin{align*} \dfrac{r + 9}{2r + 3} \times \dfrac{r + 4}{r + 4} & = \dfrac{(r + 9)(r + 4)}{(2r + 3)(r + 4)} \\ & = \dfrac{r^2 + 13r + 36}{(2r + 3)(r + 4)}\end{align*} $ Now we have: $ = \dfrac{4r^2 + 22r + 24}{(r + 4)(2r + 3)} - \dfrac{r^2 + 13r + 36}{(2r + 3)(r + 4)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4r^2 + 22r + 24 - (r^2 + 13r + 36)}{(r + 4)(2r + 3)} $ $ = \dfrac{4r^2 + 22r + 24 - r^2 - 13r - 36}{(r + 4)(2r + 3)} $ $ = \dfrac{3r^2 + 9r - 12}{(r + 4)(2r + 3)}$ Expand the denominator: $ = \dfrac{3r^2 + 9r - 12}{2r^2 + 11r + 12}$